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Is there a necessary and sufficient condition for a linearly ordered topological space to be pseudo-metrizable? (by a pseudo-metric, I mean a map $X\times X\rightarrow\mathbb{R}$ in which all the metric axioms are satisfied except that the distance between two distinct points may be zero)

A negative answer is, of course, still acceptable; I was wondering if there is an analogue of Urysohn's metrization theorem with respect to pseudo-metrics in linearly ordered spaces.

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    $\begingroup$ I don't see how a pseudo-metric could ever help you, since if $\langle X,<\rangle$ is a linear order and $d$ is a pseudo-metric on $X$ that is not a metric, then $d(x,y)=0$ for some $x\neq y$, then the topology coming from $d$ will be wrong, since every open ball containing $x$ will also contain $y$, but $(-\infty,y)$ is an open interval containing $x$ and not $y$. So the space will be pseudo-metrizable just in case it is metrizable. Or have I misunderstood the question? $\endgroup$ – Joel David Hamkins Apr 1 '16 at 20:33
  • $\begingroup$ (I had meant $x<y$.) $\endgroup$ – Joel David Hamkins Apr 1 '16 at 20:55
  • $\begingroup$ @JoelDavidHamkins Thank you! Yes, what you mean is that every topology which arises from a linear order is Hausdorff, but a topology arising from a pseudo-metric which is not a metric can never be Hausdorff, so there can be no linearly ordered space which is strictly pseudo-metric. $\endgroup$ – Simon_Peterson Apr 2 '16 at 8:24
  • $\begingroup$ @JoelDavidHamkins Why didn't you post this as an answer so I could accept it? $\endgroup$ – Simon_Peterson Apr 2 '16 at 8:24
  • $\begingroup$ I hadn't posted it as an answer, because I wasn't sure that I had understood what you wanted. But it's no problem! $\endgroup$ – Joel David Hamkins Apr 2 '16 at 12:02
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As Joel Hamkins observed above, a pseudometric may generate the topology of a linearly ordered topological space only if it is a metric.

A necessary and sufficient condition for a linearly ordered topological space $X$ to be metrizable is that is has a $G_\delta$-diagonal (that is, the set $\Delta=\{(x,x):x\in X\}$ is a $G_\delta$ subset of $X\times X$ ). See the article by Lutzer in Encyclopedia of General Topology for further references, this article is available at the web-page of the author at: http://www.resnet.wm.edu/~djlutz/Encyclopedia050602.pdf . The same article also gives a condition due to Faber for the metrizability of generalized ordered spaces (equivalently subspaces of linearly ordered topological spaces): Existence of a dense, $\sigma$-closed-discrete subset satisfying an additional condition listed there.

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